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\title{\heiti\zihao{2}习题12.4}
\author{中书君}
\date{\today}

\begin{document}
\section{求下列幂级数的收敛域}
\subsection{$\sum\limits_{n=1}^{\infty}\left(1+\dfrac{1}{n}\right)^{n^{2}}(x-1)^{n}$}
\textbf{解}\quad
由Cauchy-Handamard公式:
$$
	\varlimsup_{n\rightarrow \infty}\sqrt[n]{\left(1+\dfrac{1}{n}\right)^{n^{2}}}=\varlimsup_{n\rightarrow \infty}\left(1+\dfrac{1}{n}\right)^{n}=\mathrm{e}
$$
收敛半径$R=\dfrac{1}{\mathrm{e}}$.记$(x-1)=t$.\par
(1) $t = \dfrac{1}{\mathrm{e}}$,则级数的系数为
$$
	\dfrac{\left(1+\dfrac{1}{n}\right)^{n^{2}}}{\mathrm{e}^{n}}=\mathrm{e}^{-n^{2}\ln(1+\dfrac{1}{n})-n}\sim\mathrm{e^{-\dfrac{1}{2}}}\neq 0
$$
所以级数并不收敛.\par
(2)$t = -\dfrac{1}{\mathrm{e}}$,则级数的系数为
$$
	(-1)^{n}\dfrac{\left(1+\dfrac{1}{n}\right)^{n^{2}}}{\mathrm{e}^{n}}=(-1)^{n}\mathrm{e}^{-n^{2}\ln(1+\dfrac{1}{n})-n}\sim(-1)^{n}\mathrm{e^{-\dfrac{1}{2}}}\neq 0
$$
所以级数并不收敛.\par
综上,幂级数的收敛域为$(1-\dfrac{1}{\mathrm{e}},1+\dfrac{1}{\mathrm{e}})$
\subsection{$\sum\limits_{n=1}^{\infty}\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right) x^{n}$}
\textbf{解}\quad
由Cauchy-Handamard公式:
$$
	\varlimsup_{n\rightarrow \infty}\sqrt[n]{\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right)}=\varlimsup_{n\rightarrow \infty}\sqrt[n]{\ln n +\gamma}=1
$$
其中$\gamma$是欧拉常数.所以级数的收敛半径$R=1$.\par
$x=\pm 1$时,由于$\lim\limits_{n\rightarrow \infty}\left|1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}x^{n}\right|=\infty$,所以收敛域为$(-1, 1)$.

\subsection{$\sum\limits_{n=0}^{\infty}(-1)^{n} \dfrac{\ln (n+1)}{n+1}(x+1)^{n}$}
\textbf{解}\quad
由Cauchy-Handamard公式:
$$
	\varlimsup_{n\rightarrow \infty}\sqrt[n]{\left|(-1)^{n} \dfrac{\ln (n+1)}{n+1}\right|}=\varlimsup_{n\rightarrow \infty}\dfrac{\sqrt[n]{\ln(n+1)}}{\sqrt[n]{n+1}}=1
$$
所以收敛半径为$R=1$.记$t=x+1$\par
(1)$t=1$,由于$g'(x)=\left(\dfrac{\ln x}{x}\right)'<0(x\geqslant 3)$,所以级数为Leibniz级数,收敛.\par
(2)$t=-1$,级数系数为$\dfrac{\ln (n+1)}{n+1}$,由比较判别法:$\lim\limits_{n\rightarrow \infty}\dfrac{\dfrac{\ln(n+1)}{n+1}}{\dfrac{1}{n}}=\infty$,所以级数发散.\par
收敛域为$(-2,0]$.

\subsection{$\sum\limits_{n=0}^{\infty} \dfrac{x^{n}}{a^{\sqrt{n}}}(a>0)$}
\textbf{解}\quad
由Cauchy-Handamard公式:
$$
	\varlimsup_{n\rightarrow \infty}\sqrt[n]{\dfrac{1}{a^{\sqrt{n}}}}=\varlimsup_{n\rightarrow \infty}\dfrac{1}{a^{\dfrac{1}{\sqrt{n}}}}=1
$$
所以收敛半径为$R=1$\par
(1)$x=1$,级数系数为$\dfrac{1}{a^{\sqrt{n}}}$,当$a>1$时,由Cauchy根植判别法易知级数收敛.当$a\leqslant 1$时级数的极限不为$0$,所以级数发散.\par
(2)$x=-1$,级数系数为$(-1)^{n}\dfrac{1}{a^{\sqrt{n}}}=\dfrac{(-1)^{n}}{a^{{\sqrt{n}}}}$,当$a>1$时级数绝对收敛.当$a\leqslant 1$时,级数极限不为$0$,级数发散.\par
综上,级数收敛域当$a>1$时为$[-1,1]$,当$a\leqslant 1$时为$(-1,1)$.

\subsection{$\sum\limits_{n=1}^{\infty}(-1)^{n} \dfrac{x^{2 n}}{n \cdot 2^{n}}$}
\textbf{解}\quad
由Cauchy-Handamard公式:
$$
	\varlimsup_{n\rightarrow \infty}\sqrt[n]{\left|(-1)^{n} \dfrac{1}{n \cdot 2^{n}}\right|}=\dfrac{1}{2}
$$
所以关于$x^{2}$的级数的收敛半径为$2$,$R=\sqrt{2}$.记$t=x^{2}$\par
$t=\pm\sqrt{2}$,级数为Leibniz级数,收敛,从而级数收敛域为$[-\sqrt{2},\sqrt{2}]$.


\subsection{$\sum\limits_{n=0}^{\infty} \dfrac{(n !)^{2}}{(2 n) !} x^{n}$}
\textbf{解}\quad
由Cauchy-Handamard公式:
$$
	\varlimsup_{n\rightarrow \infty}\sqrt[n]{\dfrac{(n !)^{2}}{(2 n) !}}=\varlimsup_{n\rightarrow \infty}\sqrt[n]{\dfrac{2\pi n\left(\dfrac{n}{\mathrm{e}}\right)^{2n}}{\sqrt{4\pi n}\left(\dfrac{2n}{\mathrm{e}}\right)^{2n}}}=\varlimsup_{n\rightarrow \infty}\dfrac{\dfrac{n^{2}}{\mathrm{e}^{2}}}{\dfrac{4n^{2}}{\mathrm{e}^{2}}}=\dfrac{1}{4}
$$
所以收敛半径$R=4$.\par
当$x=4$时,有$$\varlimsup\limits_{n\rightarrow\infty}n\left(\dfrac{\dfrac{4^{n}(n!)^{2}}{(2n)!}}{\dfrac{4^{n-1}(n-1!)^{2}}{(2n-2)!}}-1\right)=\varlimsup\limits_{n\rightarrow\infty}n\left(\dfrac{4n^{2}}{2n(2n-1)}-1\right)=\varlimsup\limits_{n\rightarrow\infty}\dfrac{2n^{2}}{4n^{2}-2n}=\dfrac{1}{2}>1$$所以可由Bertrand判别法知其发散.\par
当$x=-4$时,由于$\dfrac{(n!)^{2}}{(2n)!}4^{n}\sim\dfrac{\sqrt{\pi n}}{4^{n}}4^{n}=\sqrt{\pi n}=\infty$,所以极限不为$0$,发散.\par
所以收敛域为$(-4,4)$.

\subsection{$\sum\limits_{n=1}^{\infty} \dfrac{\ln ^{2} n}{n^{n}} x^{n^{2}}$}
\textbf{解}\quad
数列为
$$a_{n}=\left\{\begin{array}{ll}
		0                       & n \neq k^{2}(k\in \mathbb{Z}) \\
		\dfrac{\ln^{2}n}{n^{n}} & n =k^{2}(k\in \mathbb{Z})
	\end{array}\right.
$$
由Cauchy-Handamard公式:
$$
	\varlimsup\limits_{n\rightarrow\infty}\sqrt[n]{|a_{n}|}=\varlimsup\limits_{n\rightarrow\infty}\sqrt[n^{2}]{\dfrac{\ln ^{2} n}{n^{n}}}=1
$$
所以收敛半径为$R=1$.\par
$x=\pm 1$时,又比较判别法:
$$
	\lim\limits_{n\rightarrow \infty}\dfrac{\dfrac{\ln^{2}n}{n^{n}}}{\dfrac{1}{n^{2}}}=0
$$
从而级数绝对收敛.\par
所以收敛域为$[-1,1]$.

\subsection{$\sum\limits_{n=1}^{\infty}\left[\dfrac{(2 n-1) ! !}{2 n ! !}\right]^{p}\left(\dfrac{x-1}{2}\right)^{n}$.}
\textbf{解}$1^{\circ}$\quad
当$p=0$时,收敛域为$(-1, 3)$.\par
当$p\neq 0$时,由于
$$
	\begin{aligned}
		\dfrac{1}{\sqrt{4n}} & = \dfrac{(\sqrt{2n}\cdot\sqrt{2n-2})(\sqrt{2n-2}\cdot\sqrt{2n-4})\cdots(\sqrt{4}\cdot\sqrt{2})}{2n!!}               \\
		                     & \leqslant\dfrac{(2n-1)!!}{2n!!}                                                                                     \\
		                     & \leqslant\dfrac{(2n-1)!!}{(\sqrt{2n+1}\cdot\sqrt{2n-1})(\sqrt{2n-1}\cdot\sqrt{2n-3})\cdots(\sqrt{3}\cdot \sqrt{1})} \\
		                     & =\dfrac{1}{\sqrt{2n+1}}
	\end{aligned}
$$
从而由夹逼准则和Cauchy-Handamard公式知:
$$
	\varlimsup\limits_{n\rightarrow\infty}\sqrt[n]{\dfrac{(2 n-1) ! !}{2 n ! !}^{p}}=1
$$
从而收敛半径$R=2$.记$t=\dfrac{x-1}{2}$\par
(1)$t=1$时,由夹逼准则知:
$$
	\dfrac{1}{\sqrt{4n}^{p}}\leqslant\left[\dfrac{(2 n-1) ! !}{2 n ! !}\right]^{p}\leqslant\dfrac{1}{\sqrt{2n+1}^{p}}
$$
显然当$p<2$时级数发散,$p\geqslant 2$时级数收敛.\par
(2)$t=-1$,由于$p>0$时,$\prod\limits_{n=1}^{\infty}\left(\dfrac{(2n-1)}{2n}\right)^{p}$每一项都小于$1$,从而$T_{n}=\prod\limits_{k=1}^{n}\left(\dfrac{(2k-1)}{2k}\right)^{p}$单调递减,且由夹逼准则可知其收敛到$0$.从而级数为Leibniz级数,所以收敛.而当$p<0$时,级数极限不为$0$,发散.\par
所以综上,级数的收敛域为
$$
	\left\{\begin{array}{ll}
		(-1,3)             & p\leqslant 0   \\
		\left[-1, 3\right) & 0<p\leqslant 2 \\
		\left[-1,3\right]  & 2<p
	\end{array}\right.
$$

\textbf{解}$2^{\circ}$\quad
$$
	\left[\dfrac{(2 n-1) ! !}{2 n ! !}\right]^{p}\sim\left(\dfrac{1}{\pi n}\right)^{p}
$$
相当于直接考虑在幂级数意义下的替换(这种情况下显然可以,因为幂级数可以只关心系数的性质)$\sum\limits_{n=1}^{\infty}\left(\dfrac{1}{\pi n}\right)^{p}\left(\dfrac{x-1}{2}\right)^{n}$.而这个级数的收敛域是很容易解出的.

\section{设正项级数 $\sum\limits_{n=1}^{\infty} a_{n}$ 发散 $, A_{n}=\sum\limits_{k=1}^{n} a_{k},$ 且 $\lim\limits_{n \rightarrow \infty} \dfrac{a_{n}}{A_{n}}=0,$ 求幂级数 $\sum\limits_{n=1}^{\infty} a_{n} x^{n}$ 和 $\sum\limits_{n=1}^{\infty} A_{n} x^{n}$ 的收敛半径.}
\textbf{解}$1$\quad
先证明二者收敛半径相等.
\begin{proof}
	由于级数$\sum\limits_{n=1}^{\infty} a_{n}$和$A_{n}>a_{n}$知$\sum\limits_{n=1}^{\infty} a_{n} x^{n}$ 和 $\sum\limits_{n=1}^{\infty} A_{n} x^{n}$当$x=1$时显然都发散,所以由Abel定理可知二者的收敛半径$R_{1}$,$R_{2}$都小于等于$1$.而且因为$A_{n}>a_{n}$,所以$R_{2}\geqslant R_{1}$\par
	又因为当$x<R_{2}$时,$\sum\limits_{n=1}^{\infty} A_{n} x^{n}$绝对收敛,从而有
	$$
		\begin{aligned}
			\sum\limits_{n=1}^{\infty} |a_{n} x^{n}| & =\sum\limits_{n=1}^{\infty} |(A_{n}-A_{n-1}) x^{n}|                                                          \\
			                                         & \leqslant\sum\limits_{n=1}^{\infty} |A_{n} x^{n}|+x\sum\limits_{n=1}^{\infty} |A_{n-1} x^{n}|\leqslant\infty
		\end{aligned}
	$$
	所以由上式知$R_{1}\geqslant R_{2}$,所以$R_{1}=R_{2}$.
\end{proof}
由于$$\lim\limits_{n \rightarrow \infty} \dfrac{a_{n}}{A_{n}}=\lim\limits_{n \rightarrow \infty}\dfrac{A_{n}-A_{n-1}}{A_{n}}=1-\dfrac{A_{n-1}}{A_{n}}=0$$所以$\lim\limits_{n \rightarrow \infty}\dfrac{A_{n-1}}{A_{n}}=1$从而$R_{1}=R_{2}=1$.\par
\textbf{解}$2$\quad
设幂级数$\sum\limits_{n=1}^{\infty} a_{n} x^{n}$ 和 $\sum\limits_{n=1}^{\infty} A_{n} x^{n}$ 的收敛半径为$R_{1},R_{2}$,由于$A_{n}>a_{n}$,所以$R_{1}\geqslant R_{2}$.显然$R_{1}\leqslant 1$.又由Stolz定理得:
$$
	\lim\limits_{n \rightarrow \infty}\dfrac{a_{n}}{A_{n}}=\lim\limits_{n \rightarrow \infty}\dfrac{a_{n}-a_{n-1}}{A_{n}-A_{n-1}}=\lim\limits_{n \rightarrow \infty}\left(1-\dfrac{a_{n-1}}{a_{n}}\right)=0
$$
所以$R_{1}=1$.\par
又因为$\lim\limits_{n \rightarrow \infty}\dfrac{A_{n}}{A_{n+1}}=\lim\limits_{n \rightarrow \infty}\dfrac{A_{n+1}-a_{n+1}}{A_{n+1}}=1$,所以$R_{2}=1$.

\section{求下列幂级数的和函数}
\subsection{$\sum\limits_{n=0}^{\infty} \dfrac{x^{2 n}}{2 n+1}$}
\textbf{解}\quad
$R=\lim \limits_{n \rightarrow \infty} \dfrac{2 n+1}{2 n+3}=1$, 当 $x=\pm 1$ 时, 级数显然发散, $\therefore$ 收敛域为 $(-1,1)$\par
$$
	(xS(x))'=\sum\limits_{n=1}^{\infty}x^{2n}=\dfrac{x^{2}}{1-x^{2}}
$$
故$S(x)=\dfrac{1}{x}\int_{0}^{x}\dfrac{t^{2}}{1-t^{2}}\mathrm{~d}t=\dfrac{1}{2x}\ln\dfrac{1+x}{1-x}-1$
\subsection{$\sum\limits_{n=1}^{\infty}(-1)^{n-1} n^{2} x^{n}$}
\textbf{解}\quad
$R=\lim\limits_{n \rightarrow \infty} \dfrac{n^{2}}{(n+1)^{2}}=1,$ 在 $x=\pm 1$ 处级数发散. $\therefore$ 收敛域为 $(-1,1)$\par
原式 $=x \sum\limits_{n=1}^{\infty}(-1)^{n-1} n^{2} x^{n-1}=x \sum\limits_{n=1}^{\infty}(-1)^{n-1} n(n+1) x^{n-1}-(-1)^{n-1} n x^{n-1}$\par
$\sum\limits_{n=1}^{\infty}(-1)^{n-1} n(n+1) x^{n-1}=\sum\limits_{n=1}^{\infty}\left[(-x)^{n+1}\right]^{\prime}=\left[\sum\limits_{n=1}^{\infty}(-x)^{n+1}\right]^{\prime \prime}=\left(\dfrac{x^{2}}{1+x}\right)^{\prime \prime}=\dfrac{2}{(1+x)^{3}}$\par
$\sum\limits_{n=1}^{\infty}(-1)^{n-1} n x^{n-1}=-\sum\limits_{n=1}^{\infty}\left[(-x)^{n}\right]^{\prime}=-\left(\dfrac{-x}{1+x}\right)^{\prime}=\dfrac{1}{(1+x)^{2}}=\dfrac{1+x}{(1+x)^{3}}$\par
$\therefore$ 原式 $=\dfrac{x(1-x)}{(1+x)^{3}}$

\subsection{$\sum\limits_{n=1}^{\infty} \dfrac{x^{n}}{n(n+1)}$}
\textbf{解}\quad
易得 $S(x)=\sum\limits_{n=1}^{\infty} \dfrac{x^{n}}{n(n+1)}$ 的收敛区间为 $[-1,1]$,所以在这个区间上级数内闭一致收敛,从而易得和函数$S(x)$在区间$[-1,1]$上连续\par
$(xS(x))''=\sum\limits_{n=0}^{\infty}x^{n}=\dfrac{1}{1-x}$,从而
$$
	S(x)=\dfrac{1}{x}\int_{0}^{x}-\ln(1-t)\mathrm{~d}t=1+\left(\dfrac{1}{x}-1\right)\ln(1-x)\qquad(x\in(-1,1))
$$
又因为$S(x)$在$[-1,1]$内连续,所以令$x\rightarrow\pm 1$就可以得到$S(1)=1,S(-1)=1-2\ln 2$\par
\textcolor{blue}{本题的思路为先算出级数的收敛半径,考虑端点发现端点处收敛,所以收敛域为$[-1,1]$,根据幂级数的基本性质可知在这个区间里内闭一致收敛,从这点可以容易导出和函数在区间$[-1,1]$内连续.当我们知道在$(-1,1)$里的$S(x)$表达式后就可以令$x\rightarrow 1^{-}$和$x\rightarrow -1^{+}$得到$S(\pm 1)$.}
\subsection{$\sum\limits_{n=1}^{\infty} n(n+1) x^{n}$}
\textbf{解}\quad
易得 $\sum\limits_{n=1}^{\infty} n(n+1) x^{n}$ 的收敛区间为 $(-1,1)$\par
$\sum\limits_{n=1}^{\infty} n(n+1) x^{n}=x \sum\limits_{n=1}^{\infty} n(n+1) x^{n-1}=x \sum\limits_{n=1}^{\infty}\left(x^{n+1}\right)^{\prime \prime}=x\left(\dfrac{x^{2}}{1-x}\right)^{\prime \prime}=\dfrac{2 x}{(1-x)^{3}}$

\subsection{$\sum\limits_{n=1}^{\infty} n^{3} x^{n}$}
\textbf{解}\quad
易得 $\sum\limits_{n=1}^{\omega} n^{3} x^{n}$ 的收敛区间为 $(-1,1)$.\par
$$\begin{aligned}\sum\limits_{n=1}^{\infty} n^{3} x^{n} & =x \sum\limits_{n=1}^{\infty} n^{3} x^{n-1}=x \sum\limits_{n=1}^{\infty}\left(n^{2} x^{n}\right)^{\prime}=x\left(\sum\limits_{n=1}^{\infty} n^{2} x^{n}\right)^{\prime}=x\left[x\left(\sum\limits_{n=1}^{\infty} n x^{n}\right)\right]^{\prime} \\
		                                       & =x\left\{x\left[x\left(x \sum\limits_{n=1}^{\infty} x^{n-1}\right)^{\prime}\right]^{\prime}\right\}^{\prime}\end{aligned}$$
$\because \sum\limits_{n=0}^{\infty} x^{n}=\dfrac{1}{1-x}$\par
$\therefore$原式 $=\dfrac{x(1+2 x)(1-x)+3 x\left(x+x^{2}\right)}{(1-x)^{4}}$

\subsection{$\sum\limits_{n=1}^{\infty} \dfrac{1}{3 !}(n+1)(n+2)(n+3) x^{n}$}
\textbf{解}\quad
易得 $\sum\limits_{n=1}^{\infty} \dfrac{1}{3 !}(n+1)(n+2)(n+3) x^{n}$ 的收敛区间为 $(-1,1)$\par
$\sum\limits_{n=1}^{\infty} \dfrac{1}{3 !}(n+1)(n+2)(n+3) x^{n}=\sum\limits_{n=1}^{\infty} \dfrac{\left(x^{n+3}\right)^{m}}{3 !}=\dfrac{1}{3 !}\left(\sum\limits_{n=1}^{\infty} x^{n+3}\right)^{\prime \prime \prime}=\dfrac{1}{6}\left(\dfrac{x^{4}}{1-x}\right)^{\prime \prime \prime}=\dfrac{1}{(1-x)^{4}}-1$

\section{证明如下等式:$$\int_{0}^{1}\left[-\dfrac{\ln (1-x)}{x}\right] \mathrm{d} x=\sum\limits_{n=1}^{\infty} \dfrac{1}{n^{2}}$$}
\begin{proof}
	由Cauchy-Handamard公式易知收敛半径$R=1$.显然当$x=\pm 1$时,由于$\sum\limits_{n=1}^{\infty}\dfrac{1}{n^{2}}$绝对收敛,从而级数收敛.所以收敛域为$[-1,1]$,而$[0,1]\subset [-1,1]$,类似3.3可知和函数$S(x)$在区间内连续.\par
	易知$(xS'(x))'=\sum\limits_{n=1}^{\infty}x^{n-1}=\dfrac{1}{1-x}\qquad(x\in(-1,1))$,所以对两侧积分即可证明
	$$
		S(1)=\lim\limits_{x\rightarrow 1^{-}}S(x)=\lim\limits_{x\rightarrow 1^{-}}\int_{0}^{x}\left[-\dfrac{\ln (1-t)}{t}\right] \mathrm{d} t=\int_{0}^{1}\left[-\dfrac{\ln (1-x)}{x}\right] \mathrm{d} x=\sum\limits_{n=1}^{\infty} \dfrac{1}{n^{2}}
	$$
\end{proof}
\textcolor{blue}{注意这里最后一步使用的是瑕积分的定义}

\section{证明:$$\dfrac{\pi}{4}=1-\dfrac{1}{3}+\dfrac{1}{5}-\cdots+\dfrac{(-1)^{n-1}}{2 n-1}+\cdots$$}
\begin{proof}
	由于级数$\sum\limits_{n=1}^{\infty}(-1)^{n-1}\dfrac{x^{2n-1}}{2n-1}$的收敛半径为$1$,收敛域为$[-1,1]$,从而和函数在区间上连续(类似3.3),所以有$S'(x)=\sum\limits_{n=0}^{\infty}(-1)^{n}x^{2n}=\dfrac{1}{1+x^{2}}$,而对$S'(x)$积分可得$S(x)=\int_{0}^{x}\dfrac{1}{1+x^{2}}\mathrm{~d}x=\arctan x$.使$x\rightarrow 1^{-}$即得$$\lim\limits_{x\rightarrow 1^{-}}\arctan x=\dfrac{\pi}{4}=1-\dfrac{1}{3}+\dfrac{1}{5}-\cdots+\dfrac{(-1)^{n-1}}{2 n-1}+\cdots$$
\end{proof}

\section{设 $f(x)=\sum\limits_{n=1}^{\infty} \dfrac{x^{n}}{n^{2} \ln (1+n)}$, 证明:}

\subsection{$f(x)$ 在 $[-1,1]$ 上连续}
\begin{proof}
	由Cauchy-Handamard公式易知幂级数收敛半径为$1$.从而在$(-1,1)$上内闭一致收敛,从而在端点处的任意空心邻域$U_{0}(1),U_{0}(-1)$内有$S_{n}x\rightrightarrows f(x)$,从而$\sum\limits_{n=1}^{\infty} \lim\limits_{x\rightarrow 1^{-}}\dfrac{x^{n}}{n^{2} \ln (1+n)}=f(1)$和$\lim\limits_{x\rightarrow 1^{-}}f(x)$都存在且相等,从而$f(x)$在$[-1,1]$上连续.
\end{proof}

\subsection{$f(x)$ 在 $x=-1$ 处可导}
\begin{proof}
	$$f'(x)=\sum\limits_{n=1}^{\infty}\dfrac{x^{n-1}}{n\ln (n+1)}(x\in[-1,1])$$
	由于
	$$
		f^{\prime}(-1)=\sum\limits_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n \ln (n+1)}
	$$
	为Leibniz级数,收敛,从而$f'(x)$的收敛域为$[-1,1)$,在这个区间上$f'(x)$内闭连续.\par
	所以 $\lim\limits_{x \rightarrow-1^{+}} f^{\prime}(x)=f_{+}^{\prime}(-1)=\sum\limits_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n \ln (n+1)}$\par
	所以 $f^{\prime}(x)$ 在 $x=-1$ 处可导.
\end{proof}



\subsection{$\lim\limits_{x \rightarrow 1^-} f^{\prime}(x)=+\infty,$ 进一步证明 $f(x)$ 在 $x=1$ 处不可导.}
\begin{proof}
	易证 $f^{\prime}(x)$ 的收敛半径为 $1,$ 从而 $f^{\prime}(x)$ 在 $(-1,1)$ 上连续, $\lim\limits_{x \rightarrow 1^{-}} f^{\prime}(x)=\sum\limits_{n=1}^{\infty} \dfrac{1}{n \ln (1+n)}$.
	显然正项级数 $\sum\limits_{n=1}^{\infty} \dfrac{1}{n \ln (1+n)}$ 发散，所以 $\lim\limits_{x \rightarrow 1^{-}} f^{\prime}(x)=+\infty$\par
	若在$x=1$处可导,则有
	$$
		\begin{aligned}
			f'(1)=\lim\limits_{x \rightarrow 1^{-}}\dfrac{f(x)-f(1)}{x-1}=\lim\limits_{x \rightarrow 1^{-}} f^{\prime}(x)=+\infty
		\end{aligned}
	$$
	(由洛必达法则)从而$\lim\limits_{x \rightarrow 1^{-}}f(x)\neq f(1)$,与$f(x)$在$[-1,1]$上连续矛盾,从而只可能是假设错误,即$f(x)$在$1$处不可导.
\end{proof}

\section{设$S(x)=\sum\limits_{n=1}^{\infty} \dfrac{x^{n}}{n^{2}}, 0 \leqslant x \leqslant 1 .$ 证明:对任意的 $x \in(0,1),$有$$\begin{array}{l}S(x)+S(1-x)+\ln x \cdot \ln (1-x) \equiv C \\\text{ 其中 }C=S(1)=\sum\limits_{n=1}^{\infty} \dfrac{1}{n^{2}}\end{array}$$}
\begin{proof}
	设 $s(x)=S(x)+S(1-x)+\ln x \cdot \ln (1-x), x \in(0,1)$
	则
	$$
		\begin{aligned}
			S^{\prime}(x) & =S^{\prime}(x)+S^{\prime}(1-x)+\dfrac{\ln (1-x)}{x}-\dfrac{\ln x}{1-x}                                                                                                                                                             \\
			              & =\sum\limits_{n=1}^{\infty} \dfrac{x^{n-1}}{n}-\sum\limits_{n=1}^{\infty} \dfrac{(1-x)^{n-1}}{n}=\dfrac{1}{x} \sum\limits_{n=1}^{\infty} \dfrac{x^{n}}{n}-\dfrac{1}{1-x} \sum\limits_{n=1}^{\infty}(-1)^{n-1} \dfrac{(x-1)^{n}}{n} \\
			              & =\sum\limits_{n=1}^{\infty} \dfrac{x^{n-1}}{n}-\sum\limits_{n=1}^{\infty} \dfrac{(1-x)^{n-1}}{n}-\sum\limits_{n=1}^{\infty} \dfrac{x^{n-1}}{n}+\dfrac{1}{1-x} \sum\limits_{n=1}^{\infty} \dfrac{(1-x)^{n}}{n}=0
		\end{aligned}
	$$
	故 $s(x)=c($ 常数 $), x \in(0,1)$\par
	又因为 $\lim \limits_{x \rightarrow 1^{-}} s(x)=S(1),$ 所以 $s(x)=S(1)$\par
	所以 $S(x)+S(1-x)+\ln x \cdot \ln (1-x) \equiv C,$ 其中 $\lim \limits_{x \rightarrow \infty} C=s(1)=\sum\limits_{n=1}^{\infty} \dfrac{1}{n^{2}}$
\end{proof}

\section{利用已知的幂级数展开,将下列函数展开为Maclaurin级数}
\subsection{$\mathrm{e}^{x^{2}}$}
\textbf{解}\quad
$$
	\mathrm{e}^{x}=\sum\limits_{n=0}^{\infty}\dfrac{x^{n}}{n!}\qquad(x\in\mathbb{R})
$$
从而
$$
	\mathrm{e}^{x^{2}}=\sum\limits_{n=0}^{\infty}\dfrac{x^{2n}}{n!}\qquad(x\in\mathbb{R})
$$
\subsection{$\cos ^{2} x$}
\textbf{解}\quad
$$
	\begin{aligned}
		\cos^{2}x & =\dfrac{1}{2}+\dfrac{\cos 2x}{2}                                                                        \\
		          & =\dfrac{1}{2}+\dfrac{\sum\limits_{n=0}^{\infty}(-1)^{n}\dfrac{(2x)^{2n}}{2n!}}{2}\qquad(x\in\mathbb{R})
	\end{aligned}
$$

\subsection{$\ln \sqrt{\dfrac{1+x}{1-x}}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\ln \sqrt{\dfrac{1+x}{1-x}} & =\dfrac{\ln(1+x)-\ln(1-x)}{2}                                                                               \\
		                            & =\dfrac{\sum\limits_{n=1}^{\infty}(-1)^{n+1}\dfrac{x^{n}}{n}+\sum\limits_{n=1}^{\infty}\dfrac{x^{n}}{n}}{2} \\
		                            & =\sum\limits_{n=1}^{\infty}\dfrac{x^{2n-1}}{2n-1}\qquad(x\in(-1,1))
	\end{aligned}
$$

\subsection{$\dfrac{x}{1+x-2 x^{2}}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\dfrac{x}{1+x-2 x^{2}} & =\dfrac{1}{3}\left(\dfrac{1}{1-x}-\dfrac{1}{1+2x}\right)                                        \\
		                       & =\dfrac{1}{3}\left(\sum\limits_{n=0}^{\infty}x^{n}-\sum\limits_{n=1}^{\infty}(-2x)^{n}\right)   \\
		                       & =\dfrac{\sum\limits_{n=1}^{\infty}(1-(-2)^{n})x^{n}}{3}\qquad(x\in(-\dfrac{1}{2},\dfrac{1}{2}))
	\end{aligned}
$$

\subsection{$(1+x) \mathrm{e}^{-x}$}
\textbf{解}\quad
$$
	\begin{aligned}
		(1+x) \mathrm{e}^{-x} & =(1+x)\left(\sum\limits_{n=0}^{\infty}\dfrac{(-x)^{n}}{n!}\right)                \\
		                      & = 1+\sum\limits_{n=1}^{\infty}\dfrac{1-n}{n!}(-1)^{n}x^{n}\qquad(x\in\mathbb{R})
	\end{aligned}
$$

\subsection{$(1+x) \ln (1+x)$}
\textbf{解}\quad
$$
	\begin{aligned}
		(1+x) \ln (1+x) & =(1+x)\left(\sum\limits_{n=1}^{\infty}\dfrac{(-1)^{n-1}x^{n}}{n}\right)      \\
		                & =x+\sum\limits_{n=2}^{\infty}\dfrac{(-1)^{n}x^{n}}{n(n-1)}\qquad(x\in(-1,1])
	\end{aligned}
$$

\subsection{$x \arctan x-\ln \sqrt{1+x^{2}}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\left(x \arctan x-\ln \sqrt{1+x^{2}}\right)' & =\arctan x                                                                                          \\
		                                             & =\sum\limits_{n=0}^{\infty}\left(\dfrac{(-1)^{n}x^{2n+1}}{2n+1}\right)\qquad(x\in\left[-1,1\right])
	\end{aligned}
$$
所以
$$
	x \arctan x-\ln \sqrt{1+x^{2}}=\sum\limits_{n=0}^{\infty}\left(\dfrac{(-1)^{n}x^{2n+2}}{(2n+2)(2n+1)}\right)\qquad(x\in\left[-1,1\right])
$$

\subsection{$\int\nolimits_{0}^{x} \dfrac{\sin t}{t} \mathrm{~d} t$}
\textbf{解}\quad
$$
	\dfrac{\sin x}{x}=\sum\limits_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2 n}}{(2 n+1) !} \qquad x \in\mathbb{R}
$$
故有
$$
	\begin{aligned}
		\int_{0}^{x} \dfrac{\sin t}{t} \mathrm{~d} t & =\int_{0}^{x} \sum\limits_{n=0}^{\infty} \dfrac{(-1)^{n}t^{2 n}}{(2 n+1) !}  \mathrm{~d} t     \\
		                                             & =\sum\limits_{0}^{\infty} \int_{0}^{x} \dfrac{(-1)^{n}t^{2 n}}{(2 n+1) !}  \mathrm{~d} t       \\
		                                             & =\sum\limits_{n=0}^{\infty} \dfrac{(-1)^{n}x^{2 n+1}}{(2 n+1) !(2 n+1)} \qquad x \in\mathbb{R}
	\end{aligned}
$$

\subsection{$\ln \left(x+\sqrt{1+x^{2}}\right)$}
\textbf{解}\quad
$$
	\begin{aligned}
		\left(\ln \left(x+\sqrt{1+x^{2}}\right)\right)' & =\dfrac{1}{\sqrt{1+x^{2}}}                                                                            \\
		                                                & =\sum\limits_{n=0}^{\infty}\left(\begin{array}{r}
				-\dfrac{1}{2} \\
				n
			\end{array}\right)x^{2n}\qquad(x\in\left[-1,1\right])
	\end{aligned}
$$
故有
$$
	\begin{aligned}
		\ln \left(x+\sqrt{1+x^{2}}\right) & =\sum\limits_{n=0}^{\infty}\left(\begin{array}{r}
				-\dfrac{1}{2} \\
				n
			\end{array}\right)\dfrac{x^{2n+1}}{2n+1}\qquad(x\in\left[-1,1\right])\qquad(x\in\left[-1,1\right])
	\end{aligned}
$$
\subsection{$\arctan \dfrac{2 x}{1-x^{2}}$}
\textbf{解}\quad
$$
	\begin{aligned}
		\left(\arctan \dfrac{2 x}{1-x^{2}}\right)' & =\dfrac{2}{1+x^{2}}                                                     \\
		                                           & =2\sum\limits_{n=0}^{\infty}(-1)^{n}x^{2n}\qquad(x\in\left(-1,1\right))
	\end{aligned}
$$
故有
$$
	\arctan \dfrac{2 x}{1-x^{2}}=2\sum\limits_{n=0}^{\infty}(-1)^{n}\dfrac{x^{2n+1}}{2n+1}\qquad(x\in\left(-1,1\right))
$$

\section{求下列函数在指定点的 Taylor 展开:}
\subsection{$1+2-3 x^{2}+5 x^{3}, \quad x_{0}=1$}
\textbf{解}\quad
$f(x)=5(x-1)^{3}+12(x-1)^{2}+11(x-1)+5\qquad x\in(-\infty,\infty)$


\subsection{$\dfrac{1}{x^{2}}, \quad x_{0}=-1$}
\textbf{解}\quad
$f^{(n)}(-1)=\dfrac{(-1)^{n}(n+1)!}{(-1)^{n+2}}=(n+1)!$,所以$f(x)=\sum\limits_{n=0}^{\infty}(n+1)(x+1)^{n}$

\subsection{$\sin x, \quad x_{0}=\dfrac{\pi}{6}$}
\textbf{解}\quad
$f^{(n)}(x)=\sin(x+\dfrac{n\pi}{2})$.将$f(x)$在$\dfrac{\pi}{6}$处展开可得:
$$
	\sin(x)=\dfrac{\sqrt{3}}{2}\sum\limits_{n=0}^{\infty}\dfrac{(-1)^{n}}{(2n+1)!}\left(x-\dfrac{\pi}{6}\right)^{2n+1}+\dfrac{1}{2}\sum\limits_{n=0}^{\infty}\dfrac{(-1)^{n}}{(2n)!}\left(x-\dfrac{\pi}{6}\right)^{2n}
$$


\subsection{$\dfrac{x-1}{x+1}, \quad x_{0}=1$.}
\textbf{解}\quad
$f^{(n)}(x)=\left(\dfrac{x-1}{x+1}\right)^{(n)}=(-1)^{n+1}\dfrac{2(n!)}{(x+1)^{n+1}}$.\par
$$
	f(x)=1+\sum\limits_{n=0}^{\infty}\dfrac{(-1)^{n+1}}{2^{n}}(x-1)^{n}\qquad x\in(-1,3)
$$

\section{应用 $\dfrac{\mathrm{e}^{x}-1}{x}$ 的 Maclaurin 展开证明:
  $$
	  \sum\limits_{n=1}^{\infty} \dfrac{n}{(n+1) !}=1
  $$}
\begin{proof}
	$\dfrac{\mathrm{e}^{x}-1}{x}=\sum\limits_{n=1}^{\infty}\dfrac{x^{n-1}}{n!}$.又因为$$\sum\limits_{n=1}^{\infty}\dfrac{n}{(n+1)!}=\sum\limits_{n=1}^{\infty}\left(\dfrac{1}{n!}-\dfrac{1}{(n+1)!}\right)$$
	所以由于$\sum\limits_{n=1}^{\infty}\dfrac{1}{n!}=\mathrm{e}^{1}-1$,$\sum\limits_{n=1}^{\infty}\dfrac{1}{(n+!)!}=\mathrm{e}^{1}-2$得
	$$
		\sum\limits_{n=1}^{\infty} \dfrac{n}{(n+1) !}=(e-1)-(e-2)=1
	$$
\end{proof}

\section{利用幂级数展开,计算 $\int_{0}^{1} \dfrac{\ln x}{1-x^{2}} \mathrm{~d} x$}
\textbf{解}\quad
$$
	\begin{aligned}
		                                                                              & \because\dfrac{1}{1-x^{2}}=\sum\limits_{n=0}^{\infty}x^{2n}                                                                                                                                          \\
		                                                                              & \therefore\int_{0}^{1} \dfrac{\ln x}{1-x^{2}} \mathrm{~d} x =\int_{0}^{1} \ln x \sum\limits_{n=0}^{\infty}x^{2n}\mathrm{~d} x =\sum\limits_{n=0}^{\infty}\int_{0}^{1} \ln x\cdot x^{2n}\mathrm{~d} x \\
		\because\sum\limits_{n=0}^{\infty}\int_{0}^{1} \ln x\cdot x^{2n}\mathrm{~d} x & =\sum\limits_{n=0}^{\infty}\int_{0}^{1} \dfrac{\ln x}{2n+1}\mathrm{~d}x^{2n+1}                                                                                                                       \\
		                                                                              & =\sum\limits_{n=0}^{\infty}\left(\left.\int_{0}^{1} \dfrac{\ln x}{2n+1}x^{2n+1}\right|_{0}^{1}-\int_{0}^{1}\dfrac{x^{2n}}{2n+1}\mathrm{~d}x\right)                                                   \\
		                                                                              & =\sum\limits_{n=0}^{\infty}-\dfrac{1}{(2n+1)^{2}}
	\end{aligned}
$$
由于
$$
	\begin{aligned}
		                                                            & \sum\limits_{n=0}^{\infty}\dfrac{1}{n^{2}}=\dfrac{\pi^{2}}{6}                                                                   \\
		                                                            & \sum\limits_{n=0}^{\infty}\dfrac{1}{(2n)^{2}}=\dfrac{\pi^{2}}{24}                                                               \\
		\therefore\sum\limits_{n=0}^{\infty}-\dfrac{1}{(2n+1)^{2}}= & \sum\limits_{n=1}^{\infty}\dfrac{1}{n^{2}}-\sum\limits_{n=1}^{\infty}\dfrac{1}{(2n)^{2}}=\dfrac{\pi^{2}}{6}-\dfrac{\pi^{2}}{24} \\
		=                                                           & \dfrac{\pi^{2}}{8}
	\end{aligned}
$$
Euler关于$\sum\limits_{n=1}^{\infty}\dfrac{1}{n^{2}}=\dfrac{\pi^{2}}{6}$(Basel问题)的证明:
\begin{proof}
	$$
		\sin x=\sum\limits_{n=0}^{\infty}(-1)^{n}\dfrac{x^{2n+1}}{(2n+1)!}
	$$
	又因为$\sin x$的根为$\pm k\pi(k\in \mathbb{Z})$,所以
	$$
		\sin x=x\prod_{n=1}^{\infty}\left(1-\dfrac{x^{2}}{n^{2}\pi^{2}}\right)
	$$
	所以考虑$x^{3}$的系数可得:
	$$
		-\sum\limits_{n=1}^{\infty}\dfrac{1}{(n\pi)^2}=-\dfrac{1}{3!}
	$$
	即
	$$
		\sum\limits_{n=1}^{\infty}\dfrac{1}{n^{2}}=\dfrac{\pi^{2}}{6}
	$$
\end{proof}
关于能将$\sin x$写为$x\prod\limits_{n=1}^{\infty}\left(1-\dfrac{x^{2}}{n^{2}\pi^{2}}\right)$的证明:
\begin{proof}
	\begin{equation}
		\sin x=x\prod\limits_{n=1}^{\infty}\left(1-\dfrac{x^{2}}{n^{2}\pi^{2}}\right)
	\end{equation}
	能够看出当$x=k\pi$时两边相等,只需要证明$x\neq k\pi $时两边也相等.\par
	可由切比雪夫多项式推得:
	$$
		\sin(2n+1)\Phi = \sin\Phi P(\sin^{2}\Phi)
	$$
	其中$P(x)$为多项式且$\deg P(x)=n$.\par
	$$
		\begin{aligned}
			 & \because\lim\limits_{\Phi\rightarrow 0}\dfrac{\sin(2n+1)\Phi}{\sin\Phi}=\lim\limits_{\Phi\rightarrow 0}P(\sin^{2}\Phi) \\
			 & \therefore P(0)=2n+1
		\end{aligned}
	$$
	又因为$\Phi_{k}=\dfrac{k\pi}{2n+1}$是左边的所有根,所以$\sin^{2}\Phi_{k}$是右侧多项式$P(x)$的所有根,所以将$x=(2n+1)\Phi$带入即得:
	$$
		\sin x=(2n+1)\sin\dfrac{x}{2n+1}\prod\limits_{k=1}^{n}\left(1-\dfrac{\sin^{2}\dfrac{x}{2n+1}}{\sin^{2}\dfrac{k\pi}{2n+1}}\right)
	$$
	再将$\sin x$分解为$U_{m}\cdot V_{m}$,其中
	$$
		\begin{aligned}
			 & U_{m}=(2n+1)\sin\dfrac{x}{2n+1}\prod\limits_{k=1}^{m}\left(1-\dfrac{\sin^{2}\dfrac{x}{2n+1}}{\sin^{2}\dfrac{k\pi}{2n+1}}\right) \\
			 & V_{m}=\prod\limits_{k=m+1}^{n}\left(1-\dfrac{\sin^{2}\dfrac{x}{2n+1}}{\sin^{2}\dfrac{k\pi}{2n+1}}\right)
		\end{aligned}
	$$
	令$n\rightarrow \infty$即得$\lim\limits_{n\rightarrow \infty}U_{m}=x\prod\limits_{k=1}^{m}\left(1-\dfrac{x^{2}}{(k\pi)^{2}}\right)$极限存在,从而 $V_{m}$极限也存在.\par
	由Jordan不等式:
	$$
		\begin{aligned}
			 & \sin^{2}\dfrac{x}{2n+1}\leqslant\dfrac{x^{2}}{(2n+1)^{2}}                                  \\
			 & \sin^{2}\dfrac{k\pi}{2n+1}\geqslant\dfrac{4}{\pi^{2}}\cdot\dfrac{k^{2}\pi^{2}}{(2n+1)^{2}}
		\end{aligned}
	$$
	从而
	$$
		1>\prod\limits_{k=m+1}^{n}\left(1-\dfrac{\sin^{2}\dfrac{x}{2n+1}}{\sin^{2}\dfrac{k\pi}{2n+1}}\right)\geqslant\prod_{k=m+1}^{n}\left(1-\dfrac{x^{2}}{4k^{2}}\right)\geqslant\prod_{k=m+1}^{\infty}\left(1-\dfrac{x^{2}}{4k^{2}}\right)
	$$
	从而
	$$
		1\geqslant \lim\limits_{n\rightarrow\infty}V_{m}\geqslant\prod_{k=m+1}^{\infty}\left(1-\dfrac{x^{2}}{4k^{2}}\right)
	$$
	又因为
	$$
		\ln\prod_{k=m+1}^{\infty}\left(1-\dfrac{x^{2}}{4k^{2}}\right)=\sum\limits_{k=m+1}^{\infty}\ln(1-\dfrac{x^{2}}{4k^{2}})\sim-\sum\limits_{k=m+1}^{\infty}\dfrac{x^{2}}{4k^{2}}<\varepsilon(n>N)
	$$
	故由Cauchy收敛准则
	$$
		\begin{aligned}
			\sin x & =\lim\limits_{m\rightarrow \infty}\lim\limits_{n\rightarrow \infty}(2n+1)\sin\dfrac{x}{2n+1}\prod\limits_{k=1}^{n}\left(1-\dfrac{\sin^{2}\dfrac{x}{2n+1}}{\sin^{2}\dfrac{k\pi}{2n+1}}\right) \\
			       & =\lim\limits_{m\rightarrow \infty}U_{m}\cdot 1                                                                                                                                               \\
			       & =x\prod\limits_{n=1}^{\infty}\left(1-\dfrac{x^{2}}{n^{2}\pi^{2}}\right)
		\end{aligned}
	$$
\end{proof}


\section{求下列函数的和函数:}
\subsection{$\sum\limits_{n=1}^{\infty} \dfrac{n^{2}+1}{2^{n} n} x^{n}$}
\textbf{解}\quad
令$t=\dfrac{x}{2}$,级数为$S(t)=\sum\limits_{n=1}^{\infty} \dfrac{n^{2}+1}{n} t^{n}=\sum\limits_{n=1}^{\infty} \left(\dfrac{1}{n}+n\right) t^{n}$,$R=2,(x\in(-2,2))$.\par
$$
	\sum\limits_{n=1}^{\infty} \left(\dfrac{1}{n}+n\right) t^{n}=\sum\limits_{n=1}^{\infty}\dfrac{t^{n}}{n}+\sum\limits_{n=1}^{\infty}n{t^{n}}=\dfrac{1}{1-t}+\dfrac{t}{(1-t)^{2}}
$$
$t=\dfrac{x}{2}\therefore\sum\limits_{n=1}^{\infty} \dfrac{n^{2}+1}{2^{n} n} x^{n}=\dfrac{4}{(2-x)^{2}}(x\in(-2,2))$

\subsection{$\sum\limits_{n=1}^{\infty} \dfrac{(-1)^{n-1}}{n(n+1)}\left(\dfrac{2+x}{2-x}\right)^{n}$}
\textbf{解}\quad
令$t=-\dfrac{2+x}{2-x}$,级数为
$$
	S(t)=-\sum\limits_{n=1}^{\infty}\dfrac{t^{n}}{n(n+1)}(t\in[-1,1])
$$
所以
$$
	tS(t)=-\sum\limits_{n=1}^{\infty}\dfrac{t^{n+1}}{n(n+1)}
$$
从而
$$
	(tS(t))''=-\sum\limits_{n=0}^{\infty}t^{n}=\dfrac{1}{1-t}
$$
故$S(t)=\dfrac{\ln(1-t)}{t}-t\ln(1-t)-1+t$.将$t=\left(\dfrac{2+x}{2-x}\right)$带入即可$(x\in(-\infty,0]$.


\subsection{$\sum\limits_{n=1}^{\infty}\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right) x^{n}$}
\textbf{解}\quad
显然其为Cauchy乘积的形式,从而考虑$\sum\limits_{n=1}^{\infty}\dfrac{x^{n}}{n}$与$\sum\limits_{n=0}^{\infty}x^{n}$的Cauchy乘积即可$x\in(-1,1)$.
$$
	\begin{aligned}
		 & \sum\limits_{n=1}^{\infty}\dfrac{x^{n}}{n}=-\ln(1-x) \\
		 & \sum\limits_{n=1}^{\infty}x^{n}=\dfrac{1}{1-x}
	\end{aligned}
$$
故
$$
	\sum\limits_{n=1}^{\infty}\left(1+\dfrac{1}{2}+\cdots+\dfrac{1}{n}\right) x^{n}=\dfrac{-\ln(1-x)}{(1-x)}
$$

\section{求下列级数的和}
\subsection{$\sum\limits_{n=2}^{\infty} \dfrac{1}{2^{n}\left(n^{2}-1\right)}$}
\textbf{解}\quad
对于级数
$$
	\sum\limits_{n=2}^{\infty} \dfrac{x^{n}}{\left(n^{2}-1\right)}
$$
收敛域为$[-1,1]$.
$$
	\begin{aligned}
		\sum\limits_{n=2}^{\infty} \dfrac{x^{n}}{\left(n^{2}-1\right)} & =\sum\limits_{n=2}^{\infty}\dfrac{1}{2}\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)x^{n}=\dfrac{1}{2}\left(\sum\limits_{n=2}^{\infty}\dfrac{x^{n}}{n-1}-\sum\limits_{n=2}^{\infty}\dfrac{x^{n}}{n+1}\right) \\
		                                                               & =\dfrac{1}{2}\left(-x\ln(1-x)+\dfrac{(1-x)^{2}}{2x}-\dfrac{1}{2x}+2+\dfrac{\ln(1-x)}{x}\right)
	\end{aligned}
$$
当$x=\dfrac{1}{2}$时和为$\dfrac{5}{8}-\dfrac{3}{4}\ln 2$


\subsection{$\sum\limits_{n=1}^{\infty}(-1)^{n} \dfrac{n}{2^{n}}$}
\textbf{解}\quad
$$
	\sum\limits_{n=1}^{\infty}nx^{n}=\dfrac{x}{(1-x)^{2}}
$$
收敛半径为$1$.当$x=-\dfrac{1}{2}$时有
$$
	\sum\limits_{n=1}^{\infty}(-1)^{n} \dfrac{n}{2^{n}}=-\dfrac{2}{9}
$$
\subsection{$\sum\limits_{n=1}^{\infty} \dfrac{n(n+2)}{4^{n+1}}$}
\textbf{解}\quad
设 $f(x)=\sum\limits_{n=1}^{\infty}(n+2) n x^{n+1},$ 利用逐项求导公式可得
$$
	f(x)=\sum\limits_{n=1}^{\infty}(n+2) n x^{n+1}=\sum\limits_{n=1}^{\infty} n\left(x^{n+2}\right)^{\prime}=\left(\sum\limits_{n=1}^{\infty} n x^{n+2}\right)^{\prime}
$$
从而
$$
	\int_{0}^{x} f(t) \mathrm{d} t=\sum\limits_{n=1}^{\infty} n x^{n+2}=x^{3} \sum\limits_{n=1}^{\infty} n x^{n-1}=x^{3}\left(\dfrac{x}{1-x}\right)^{\prime}=\dfrac{x^{3}}{(1-x)^{2}}
$$
可得 $f(x)=\dfrac{x^{2}(3-x)}{(1-x)^{3}}$,代人 $x=\dfrac{1}{4},$ 有
$$
	f\left(\dfrac{1}{4}\right)=\dfrac{\left(3-\dfrac{1}{4}\right)\left(\dfrac{1}{4}\right)^{2}}{\left(1-\dfrac{1}{4}\right)^{3}}=\dfrac{11}{27}
$$

\subsection{$\sum\limits_{n=1}^{\infty} \dfrac{(n+1)^{2}}{2^{n}}$}
\textbf{解}\quad
设 $S(x)=\sum\limits_{n=1}^{\infty}(n+1)^{2} x^{n},$ 逐项积分可得
$$
	\int_{0}^{x} f(t) \mathrm{d} t=\sum\limits_{n=1}^{\infty}(n+1) x^{n+1}=\sum\limits_{n=2}^{\infty} n x^{n}=\dfrac{x}{(1-x)^{2}}-x
$$
所以 $f(x)=\dfrac{x\left(4-3 x+x^{2}\right)}{(1-x)^{3}}, \sum\limits_{n=0}^{\infty}(n+1)^{2}\left(\dfrac{1}{2}\right)^{n}=f\left(\dfrac{1}{2}\right)=11$

\subsection{$\sum\limits_{n=1}^{\infty}(-1)^{n} \dfrac{2^{n+1}}{n !} \cdot x, y \geqslant 0$}
\textbf{解}\quad
设 $f(x)=\sum\limits_{n=1}^{\infty} \dfrac{(-1)^{n}}{n !} x^{n+1},$ 则 $f(x)=x \sum\limits_{n=1}^{\infty} \dfrac{(-1)^{n}}{n !} x^{n}=x\left(\mathrm{e}^{-x}-1\right)$\par
$\sum\limits_{n=1}^{\infty} \dfrac{(-1)^{n}}{n !} 2^{n+1}=f(2)=2 \mathrm{e}^{-2}-2=\dfrac{2}{\mathrm{e}^{2}}-2$

\section{设曲线 $x^{\dfrac{1}{n}}+y^{\dfrac{1}{n}}=1(n>1)$ 在第一象限与坐标轴围成的面积为 $I_{n},$ 证明 :}
\subsection{$I_{n}=2 n \int_{0}^{1}\left(1-t^{2}\right)^{n} t^{2 n-1} \mathrm{~d} t$}
\begin{proof}
	由已知条件 $$I_{n}=\int_{0}^{1} y(x) \mathrm{d} x=\int_{0}^{1}(1-\sqrt[n]{x})^{n} \mathrm{~d} x .$$
	令 $x=t^{2 n},$ 则 $\mathrm{d} x=2 n t^{2 n-1} \mathrm{~d} t,$ 从而
	$$I_{n}=\int_{0}^{1} y(x) \mathrm{d} x=\int_{0}^{1}\left(1-t^{2}\right)^{n} 2 n t^{2 n-1} \mathrm{~d} t=2 n \int_{0}^{1}\left(1-t^{2}\right)^{n} t^{2 n-1} \mathrm{~d} t$$
\end{proof}
\subsection{$\sum\limits_{n=1}^{\infty} I_{n}<4$}
\begin{proof}
	$$
		\begin{aligned} I_{n} & =2 n \int_{0}^{1}\left(1-t^{2}\right)^{n} t^{2 n-1} \mathrm{~d} t=2 n \int_{0}^{1}\left(1-t^{2}\right)^{n-1}\left(1-t^{2}\right) t^{2 n-2} t \mathrm{~d} t \\ & \leqslant 2 n \int_{0}^{1}\left(1-t^{2}\right)^{n-1} t^{2 n-2} \mathrm{~d} t=2 n \int_{0}^{1}\left[\left(1-t^{2}\right) t^{2}\right]^{n-1} \mathrm{~d} t \\ & \leqslant 2 n \int_{0}^{1}\left(\dfrac{1}{4}\right)^{n-1} \mathrm{~d} t=\dfrac{2 n}{4^{n-1}}
		\end{aligned}
	$$
	另一方面，已知 $\sum\limits_{n=1}^{\infty} n x^{n-1}=\dfrac{1}{(1-x)^{2}},$ 所以 $\sum\limits_{n=1}^{\infty} n\left(\dfrac{1}{4}\right)^{n-1}=\dfrac{1}{\left(1-\dfrac{1}{4}\right)^{2}}=\dfrac{16}{9},$ 所以
	$\sum\limits_{n=0}^{\infty} I_{n} \leqslant 2 \sum\limits_{n=0}^{\infty} n\left(\dfrac{1}{4}\right)^{n-1}=2 \cdot \dfrac{16}{9}<4$
\end{proof}



\end{document}